Respuesta :
We can visualize the problem in another way, which is equivalent but easier to solve: let's imagine we hold one proton in the same place, and we move the other proton from a distance of 2.00×10−10 m to a distance of 3.00×10−15 m from the first proton. How much work is done?
The work done is equal to the electric potential energy gained by the proton:
[tex]W=q \Delta V[/tex]
where [tex]q=1.6 \cdot 10^{-19}C[/tex] is the charge of the proton and [tex]\Delta V[/tex] is the potential difference between the final position and the initial position of the proton. To calculate this [tex]\Delta V[/tex], we must calculate the electric potential generated by the proton at rest at the two points, using the formula:
[tex]V=k\frac{Q}{r}[/tex]
where [tex]k=9.0 \cdot 10^9 N m^2 C^{-2}[/tex] is the Coulomb constant and Q is the proton charge. Substituting the initial and final distance of the second proton, we find
[tex]V_i = (9.0 \cdot 10^9 )\frac{1.6 \cdot 10^{-19}}{2.0 \cdot 10^{-10}}=7.2 V[/tex]
[tex]V_f = (9.0 \cdot 10^9 )\frac{1.6 \cdot 10^{-19}}{3.0 \cdot 10^{-15}}=4.8 \cdot 10^5 V[/tex]
Therefore, the work done is
[tex]W=q \Delta V=(1.6 \cdot 10^{-19}C)(4.8 \cdot 10^5 V-72 V)=7.7 \cdot 10^{-14} J[/tex]
Electric potential is the amount of work needed to move a unit charge from a point to a specific point against an electric field. The work would it take to push two protons will be 7.7×10⁻¹⁴.
What is electric potential?
Electric potential is the amount of work needed to move a unit charge from a point to a specific point against an electric field.
The given data in the problem is;
q is the charge= 1.6 ×10⁻¹⁹ C
V is the electric potential
r₁ is the first separation distance= 2.00×10−10 m
r₂ is the second separation distance= 3.00×10−15 m
The electric potential generated by the proton at rest at the two points, using the formula:
Firstly the electric potential at loction 1
[tex]\rm V=\frac{Kq}{r} \\\\ v_i= 9\times 10^9 \times \frac{1.6\times10^{-19}}{2.0\times10^{-10}}[/tex]
The electric potential at loction 2
[tex]V_f = 9 \times 10^9 \frac{1.6 \times 10^{-19}}{3.0\times10^{-15}} \\\\ \rm v_f= 4.8 \times10^5 \ V[/tex]
The product of difference of electric potential and charge is defined as the work done.
[tex]\rm W= q \triangle V \\\\ \rm W= 1.6 \times 10^-19 \times( 4.8\times10^5 -7.2) \\\\ \rm W= 7.7 \times 10^{-14}[/tex]
Hence the work would it take to push two protons will be 7.7×10⁻¹⁴.
To learn more about electric potential work refer to the link.
https://brainly.com/question/12707371
